`data.table`全局搜索 - 過濾器行在`any`列給出模式匹配

Question

是否有任何已有的便利函數可以過濾data.table中的行，給定搜索模式，在所有列中查找？

names(DT) 

[1] "Name" "LongName" "SomeOtherCol" "NumericCol" "bar" "foo" 

這樣的事情，概括爲任意數量的列：

DT[Name %like% pattern | LongName %like% pattern | SomeOtherCol %like% pattern | bar %like% pattern | foo %like% pattern] 

Answer 1

一種方法是循環訪問列，應用您的正則表達式，這將返回邏輯data.table。您可以使用rowSums來獲取行。

dt <- data.table(a=c("Aa1","bb","1c"),b=c("A1","a1","1C"), c=letters[1:3]) 
# "a1" is the pattern to search for 
ldt <- dt[, lapply(.SD, function(x) grepl("a1", x, perl=TRUE))] 
dt[rowSums(ldt)>0] 
#  a b c 
# 1: Aa1 A1 a 
# 2: bb a1 b 

Answer 2

我不賭，這是做到這一點的最好辦法。但它服務的宗旨：

> dt <- data.table(a=c("a1","bb","1c"),b=c("A1","BB","1C")) 
> dt 
    a b 
1: a1 A1 
2: bb BB 
3: 1c 1C 

> combined <- apply(dt,1,function(r) paste(r,collapse="/%/")) 
> combined 
[1] "a1/%/A1" "bb/%/BB" "1c/%/1C" 

> grepped <- grepl("[a-z][0-9]",apply(dt,1,function(r) paste(r,collapse="/"))) 
> grepped 
[1] TRUE FALSE FALSE 

> dt[grepped,] 
    a b 
1: a1 A1 

的「/％/」必須是東西是不相關的模式和可靠地分離列。

這些步驟可以合併爲單個表達式。

Answer 3

解決方案3：

首先構建邏輯grepexpression附加的所有列。然後eval一氣呵成的整體表現：

dt <- data.table(a=c("a1","bb","1c"),b=c("A1","BB","1C")) 

search.data.table <- function(x, pattern) { 
    nms <- names(x) 
    string <- eval(expression(paste0("grepl('", 
            pattern, 
            "', ", 
            nms,", 
            ignore.case=TRUE, perl=FALSE)", 
            collapse = " | "))) 
    x[eval(as.call(parse(text=string))[[1]])] 
} 

search.data.table(dt, "a1") 
#  a b c 
# 1: Aa1 A1 a 
# 2: bb a1 b 

標杆

# functions 

Raffael <- function(x, pattern) { 
# unfortunately this implementation throws an error so I can't run the benchmark test. 
# Any help? 
    combined <- apply(x,1,function(r) paste(r,collapse="/%/")) 
    grepped <- grepl(pattern,apply(x,1,function(r) paste(r,collapse="/"))) 
    x[grepped,] 
} 

Arun <- function(x, pattern) { 
    ldt <- x[, lapply(.SD, function(x) grepl(pattern, x, perl=TRUE, ignore.case=TRUE))] 
    x[rowSums(ldt)>0] 
} 

DanielKrizian <- function(x, pattern) { 
    nms <- names(x) 
    string <- eval(expression(paste0("grepl('", pattern, "', ",nms,", ignore.case=TRUE,  perl=FALSE)",collapse = " | "))) 
    x[eval(as.call(parse(text=string))[[1]])] 
} 

# generate 1000 x 1000 benchmark data.table 

require(data.table) 
expr <- quote(paste0(sample(c(LETTERS,tolower(LETTERS),0:9),12, replace=T) 
       ,collapse="")) 
set.seed(1) 
BIGISH <- data.table(matrix(replicate(1000*1000,eval(expr)),nrow = 1000)) 
object.size(BIGISH) # 68520912 bytes 

# test 

benchmark(
    DK <- DanielKrizian(BIGISH,"qx"), 
    A <- Arun(BIGISH,"qx"), 
    replications=100) 

結果

       test replications elapsed relative user.self sys.self user.child sys.child 
2   A <- Arun(BIGISH, "qx")   100 57.72 1.000  51.95  0.44   NA  NA 
1 DK <- DanielKrizian(BIGISH, "qx")   100 59.28 1.027  53.72  0.50   NA  NA 

identical(DK,A) 
[1] TRUE